Diagnosing Hypotheses in Fubini-Tonelli Problems

A Checklist Before Exchanging Orders

A formula involving a sum and an integral invites an exchange of order, but the notation alone does not justify one. The relevant questions are whether the function is measurable on the product space, whether it is nonnegative or absolutely integrable, and whether the measure spaces satisfy the sigma-finiteness hypothesis of the standard Fubini–Tonelli theorems. Checking these conditions first distinguishes a valid exchange from one that merely produces plausible-looking iterated answers.

For a sequence \((f_n)\) on a measure space \((X,\mathcal{M},\mu)\), a mixed sum and integral can be viewed on \(\mathbb{N}\times X\). Give \(\mathbb{N}=\{1,2,\ldots\}\) its full power-set sigma-algebra and counting measure \(\#\). Then summing over \(n\) is integration with respect to \(\#\). The product measure space is \((\mathbb{N}\times X,\mathcal{P}(\mathbb{N})\otimes\mathcal{M},\#\times\mu)\). Counting measure is sigma-finite, since \(\mathbb{N}\) is a countable union of singletons.

Measurability check: If each \(f_n:X\to\mathbb{R}\) is measurable, then \(F(n,x)=f_n(x)\) is measurable on \(\mathbb{N}\times X\). Indeed, for any real \(a\), \(\{(n,x):F(n,x)>a\}=\bigcup_{n\ge1}\{n\}\times\{x:f_n(x)>a\}\), a countable union of product-measurable sets. For general product spaces, measurability of the sections alone is not the product-measurability hypothesis; check that the function itself is measurable for the product sigma-algebra.

Two standard theorems govern the exchange. Tonelli’s theorem, established in the earlier material on product integration, applies to a nonnegative product-measurable function and sigma-finite measure spaces; the common value may be \(+\infty\). Fubini’s theorem, also established earlier, applies to a real- or complex-valued product-measurable function whose absolute value has finite product integral. In that case the sections are integrable almost everywhere, and the iterated integrals equal the product integral and are finite. “Integrable” here means that the integral of the absolute value is finite; merely having finite-looking section integrals is not a substitute.

Diagnostic rule: Use Tonelli when the function is nonnegative and product-measurable, with sigma-finite factor spaces; no finiteness of the resulting integral is required. Use Fubini for a signed or complex-valued function when it is product-measurable and \(\int |F|\,d(\mu\times\nu)<\infty\), again with sigma-finite factor spaces. If neither set of conditions holds, these theorems do not license an exchange.

Mixed Sums and Integrals Under Tonelli

Worked Example: A Nonnegative Series of Integrals

Let \(X=[0,1]\) with Lebesgue measure \(m\), and for \(n\ge1\) set \(f_n(x)=2^{-n}\chi_{[0,1]}(x)\). Compute the sum of the integrals and the integral of the sum, verifying the hypotheses for the exchange.

Each \(f_n\) is nonnegative and Lebesgue measurable. The associated function \(F(n,x)=f_n(x)\) is product-measurable by the measurability check above. The factor spaces are sigma-finite: \(\mathbb{N}\) is a countable union of finite-measure singletons, and \([0,1]\) has measure \(1\). Thus Tonelli applies, including if either side is infinite. Here each term has integral

\int_{[0,1]} f_n\,dm =2^{-n}m([0,1]) =2^{-n}\cdot1 =2^{-n}.

Consequently, summing the integrals gives \(\sum_{n=1}^{\infty}2^{-n}=1\), by the geometric-series formula. At each \(x\in[0,1]\), including both endpoints, \(\sum_{n=1}^{\infty}f_n(x)=\sum_{n=1}^{\infty}2^{-n}=1\). Therefore \(\int_{[0,1]}\sum_n f_n\,dm=1\cdot m([0,1])=1\). Check: both orders give \(1\); the pointwise sum is the constant function \(1\) on a set of measure \(1\).

The example uses nonnegativity, not an unproved convergence of integrals: Tonelli itself justifies the exchange. It permits an infinite answer as well. In contrast, for a signed function the positive and negative contributions could each be infinite, so that cancellation would not define an integral. The next example uses absolute integrability to control cancellation.

Signed Terms: Check Absolute Integrability

Worked Example: An Absolutely Integrable Alternating Series

On \([0,1]\) with Lebesgue measure \(m\), define \(h_n(x)=(-1)^n2^{-n}\chi_{[0,1]}(x)\), for \(n\ge1\). Compute both orders of integration and check the absolute-integrability condition.

The function \(H(n,x)=h_n(x)\) is product-measurable, and the factor spaces are sigma-finite as in the previous example. Its absolute value is \(|h_n(x)|=2^{-n}\chi_{[0,1]}(x)\). Applying Tonelli to this nonnegative function gives

\int_{\mathbb{N}\times[0,1]}|H|\,d(\#\times m) =\sum_{n=1}^{\infty}\int_{[0,1]}2^{-n}\chi_{[0,1]}\,dm =\sum_{n=1}^{\infty}2^{-n} =1.

Thus \(H\) is absolutely integrable, so Fubini applies. For each \(n\), \(\int_{[0,1]}h_n\,dm=(-1)^n2^{-n}\cdot1=(-1)^n2^{-n}\). The resulting series is geometric: \(\sum_{n=1}^{\infty}(-1/2)^n=(-1/2)/(1-(-1/2))=-1/3\). In the other order, for every \(x\in[0,1]\), the sum is \(\sum_{n=1}^{\infty}(-1/2)^n=-1/3\), whose integral over \([0,1]\) is \((-1/3)\cdot1=-1/3\). Check: each order gives \(-1/3\), and the absolute integral is \(1<\infty\), as required by Fubini.

Why absolute integrability matters: For a signed or complex-valued function, the condition is \(\int |F|\,d(\mu\times\nu)<\infty\), not simply that the individual iterated integrals can be assigned values. Absolute integrability prevents infinite positive and negative contributions from being rearranged into different results.

Diagnosing a Failed Exchange

Worked Example: Different Iterated Sums Without Absolute Integrability

On \(\mathbb{N}\times\mathbb{N}\) with counting measure in both coordinates, define \(b_{mn}=1\) when \(n=m\), \(b_{mn}=-1\) when \(n=m+1\), and \(b_{mn}=0\) otherwise. Compute the two iterated sums and test whether Fubini applies.

Every function on this countable space is measurable. Both counting-measure spaces are sigma-finite: each is covered by its singleton sets. For a fixed \(m\), the only nonzero terms in the row are \(b_{m,m}=1\) and \(b_{m,m+1}=-1\). Thus \(\sum_{n\ge1}b_{mn}=1-1=0\), and summing these row sums over \(m\) gives \(0\).

For a fixed column \(n=1\), the only nonzero term is \(b_{1,1}=1\), so its column sum is \(1\). For each \(n\ge2\), the nonzero terms are \(b_{n,n}=1\) and \(b_{n-1,n}=-1\), so that column sum is \(1-1=0\). Summing the column sums over \(n\) therefore gives \(1+0+0+\cdots=1\). The two iterated sums differ.

To check absolute integrability, each row contains two terms of absolute value \(1\), so \(\sum_{m,n}|b_{mn}|\ge\sum_{m=1}^{\infty}2=\infty\). Thus the function is not integrable in absolute value, and Fubini does not apply. Check: the row-first answer is \(0\), while the column-first answer is \(1\); the missing condition is absolute integrability, not measurability or sigma-finiteness.

Common pitfall. The incorrect calculation is: “The array is measurable and both factor measures are sigma-finite, so Fubini says the two iterated sums are equal.” The correction is to check all of Fubini’s hypotheses. For this array, \(\sum_{m,n}|b_{mn}|=\infty\), so measurability and sigma-finiteness alone are insufficient. The computed values \(0\) and \(1\) directly confirm that the exchange is invalid.

What Sigma-Finiteness Does in the Checklist

Sigma-finiteness means that the space is a countable union of measurable sets of finite measure. It is distinct from finiteness: a finite measure space is sigma-finite by taking the whole space as one set, while a sigma-finite space may have infinite total measure. The standard product-integration theorems used here assume sigma-finiteness of the factor spaces. If that hypothesis is missing, the theorem in this form cannot be invoked; this does not by itself prove that a particular exchange is false. A specific problem may still be settled by a direct calculation or another applicable result.

For a mixed sum and integral over \(\mathbb{N}\times X\), the counting-measure factor is automatically sigma-finite. The other factor still needs checking. For instance, Lebesgue measure on \(\mathbb{R}\) is sigma-finite because \(\mathbb{R}=\bigcup_{k\ge1}[-k,k]\), and \(m([-k,k])=2k<\infty\) for each \(k\). Counting measure on an uncountable set is not sigma-finite: any finite-measure set is finite, and a countable union of finite sets is at most countable. In such a case, do not cite the standard sigma-finite product theorem without further justification.

A reliable calculation therefore proceeds in order: identify the product sigma-algebra and verify product measurability; check sigma-finiteness of both factors; then choose Tonelli for nonnegative functions or Fubini for functions with finite absolute integral. If a required condition is absent, stop before exchanging orders and determine whether a separate argument is available.

Check Your Understanding

For each question, state which hypotheses you would check before exchanging the orders of summation and integration.

For measurable \(f_n\ge0\) on a sigma-finite measure space, which theorem permits \(\int\sum_n f_n\,d\mu=\sum_n\int f_n\,d\mu\), and may the common value be infinite?
Why does measurability of each \(f_n:X\to\mathbb{R}\) imply product measurability of \(F(n,x)=f_n(x)\) on \(\mathbb{N}\times X\)?
For a signed product-measurable function, what condition on its absolute value permits Fubini’s theorem?
In the array \(b_{mn}\) above, compute the row sums and column sums and identify the failed Fubini hypothesis.
Does the failure of sigma-finiteness alone show that two iterated integrals differ? Explain what it does imply about invoking the standard sigma-finite product-integration theorems.